Question

A hyperbola having the transverse axis of length $2\, sin\, \theta$, is confocal with the ellipse $3x^2 + 4y^2 = 12$. Then its equation is

• A. $ x^{2}\, cosec^{2} \,\theta - y^{2}\, sec^{2} \,\theta = 1$
• B. $ x^{2}\, sec^{2} \,\theta - y^{2}\,cosec^{2} \,\theta = 1$
• C. $ x^{2}\, sin^{2} \,\theta - y^{2}\,cos^{2} \,\theta = 1$
• D. $ x^{2}\, cos^{2} \,\theta - y^{2}\,sin^{2} \,\theta = 1$

Answer 1

The Correct Option is A

Equation of the ellipse is $3x^2 + 4y^2 = 12$ $\Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{3} = 1 \quad.... \left(1\right)$ Eccentricity $e_{1} = \sqrt{1-\frac{3}{4}} = \frac{1}{2}$ So, the foci of ellipse are $\left(1, 0\right)$ and $\left(- 1, 0\right)$ Let the equation of the required hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1 \quad .... \left(2\right)$ Given $2a = 2\,sin \,\theta \Rightarrow a =\,sin \,\theta$ Since the ellipse $\left(1\right)$ and the hyperbola $\left(2\right)$ are confocal, so the foci of hyperbola are $\left(1, 0\right)$ and $\left(- 1, 0\right)$ too. If the eccentricity, of hyperbola be $e_{2}$ then $ae_{2} = 1 \Rightarrow \,sin \,\theta \, e_{2} = 1 \Rightarrow e_{2} = cosec\, \theta$ $\therefore\quad b^{2} = a^{2} \left(e^{2}_{2} - 1\right) = sin^{2}\,\theta \left(cosec^{2} \,\theta - 1\right) = cos^{2} \,\theta $ $\therefore \quad$ Required equation of the hyperbola is $\frac{x^{2}}{sin^{2}\,\theta } - \frac{y^{2}}{cos^{2} \,\theta } = 1 \Rightarrow x^{2}\, cosec^{2} \,\theta - y^{2}\, sec^{2} \,\theta = 1$